Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{k^2 - 12k + 32}{2k + 20} \div \dfrac{k - 8}{k + 10} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{k^2 - 12k + 32}{2k + 20} \times \dfrac{k + 10}{k - 8} $ First factor the quadratic. $n = \dfrac{(k - 8)(k - 4)}{2k + 20} \times \dfrac{k + 10}{k - 8} $ Then factor out any other terms. $n = \dfrac{(k - 8)(k - 4)}{2(k + 10)} \times \dfrac{k + 10}{k - 8} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (k - 8)(k - 4) \times (k + 10) } { 2(k + 10) \times (k - 8) } $ $n = \dfrac{ (k - 8)(k - 4)(k + 10)}{ 2(k + 10)(k - 8)} $ Notice that $(k + 10)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(k - 8)}(k - 4)(k + 10)}{ 2(k + 10)\cancel{(k - 8)}} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $n = \dfrac{ \cancel{(k - 8)}(k - 4)\cancel{(k + 10)}}{ 2\cancel{(k + 10)}\cancel{(k - 8)}} $ We are dividing by $k + 10$ , so $k + 10 \neq 0$ Therefore, $k \neq -10$ $n = \dfrac{k - 4}{2} ; \space k \neq 8 ; \space k \neq -10 $